Rearrange array such that arr[i] >= arr[j] if i is even and arr[i]<=arr[j] if i is odd and j < i
Given an array of n elements. Our task is to write a program to rearrange the array such that elements at even positions are greater than all elements before it and elements at odd positions are less than all elements before it.
Examples:
Input : arr[] = {1, 2, 3, 4, 5, 6, 7} Output : 4 5 3 6 2 7 1 Input : arr[] = {1, 2, 1, 4, 5, 6, 8, 8} Output : 4 5 2 6 1 8 1 8
The idea to solve this problem is to first create an auxiliary copy of the original array and sort the copied array. Now total number of even position in array with n elements will be floor(n/2) and remaining is the number of odd positions. Now fill the odd and even positions in the original array using the sorted array in the below manner:
- Total odd positions will be n – floor(n/2). Start from (n-floor(n/2))th position in the sorted array and copy the element to 1st position of sorted array. Start traversing the sorted array from this position towards left and keep filling the odd positions in the original array towards right.
- Start traversing the sorted array starting from (n-floor(n/2)+1)th position towards right and keep filling the original array starting from 2nd position.
Below is the implementation of above idea:
C++
// C++ program to rearrange the array // as per the given condition #include <bits/stdc++.h> using namespace std; // function to rearrange the array void rearrangeArr( int arr[], int n) { // total even positions int evenPos = n / 2; // total odd positions int oddPos = n - evenPos; int tempArr[n]; // copy original array in an // auxiliary array for ( int i = 0; i < n; i++) tempArr[i] = arr[i]; // sort the auxiliary array sort(tempArr, tempArr + n); int j = oddPos - 1; // fill up odd position in original // array for ( int i = 0; i < n; i += 2) { arr[i] = tempArr[j]; j--; } j = oddPos; // fill up even positions in original // array for ( int i = 1; i < n; i += 2) { arr[i] = tempArr[j]; j++; } // display array for ( int i = 0; i < n; i++) cout << arr[i] << " " ; } // Driver code int main() { int arr[] = { 1, 2, 3, 4, 5, 6, 7 }; int size = sizeof (arr) / sizeof (arr[0]); rearrangeArr(arr, size); return 0; } |
Java
// Java program to rearrange the array // as per the given condition import java.util.*; import java.lang.*; public class GfG{ // function to rearrange the array public static void rearrangeArr( int arr[], int n) { // total even positions int evenPos = n / 2 ; // total odd positions int oddPos = n - evenPos; int [] tempArr = new int [n]; // copy original array in an // auxiliary array for ( int i = 0 ; i < n; i++) tempArr[i] = arr[i]; // sort the auxiliary array Arrays.sort(tempArr); int j = oddPos - 1 ; // fill up odd position in // original array for ( int i = 0 ; i < n; i += 2 ) { arr[i] = tempArr[j]; j--; } j = oddPos; // fill up even positions in // original array for ( int i = 1 ; i < n; i += 2 ) { arr[i] = tempArr[j]; j++; } // display array for ( int i = 0 ; i < n; i++) System.out.print(arr[i] + " " ); } // Driver function public static void main(String argc[]){ int [] arr = new int []{ 1 , 2 , 3 , 4 , 5 , 6 , 7 }; int size = 7 ; rearrangeArr(arr, size); } } /* This code is contributed by Sagar Shukla */ |
Python3
# Python3 code to rearrange the array # as per the given condition import array as a import numpy as np # function to rearrange the array def rearrangeArr(arr, n): # total even positions evenPos = int (n / 2 ) # total odd positions oddPos = n - evenPos # initialising empty array in python tempArr = np.empty(n, dtype = object ) # copy original array in an # auxiliary array for i in range ( 0 , n): tempArr[i] = arr[i] # sort the auxiliary array tempArr.sort() j = oddPos - 1 # fill up odd position in original # array for i in range ( 0 , n, 2 ): arr[i] = tempArr[j] j = j - 1 j = oddPos # fill up even positions in original # array for i in range ( 1 , n, 2 ): arr[i] = tempArr[j] j = j + 1 # display array for i in range ( 0 , n): print (arr[i], end = ' ' ) # Driver code arr = a.array( 'i' , [ 1 , 2 , 3 , 4 , 5 , 6 , 7 ]) rearrangeArr(arr, 7 ) # This code is contributed by saloni1297 |
C#
// C# program to rearrange the array // as per the given condition using System; public class GfG { // Function to rearrange the array public static void rearrangeArr( int []arr, int n) { // total even positions int evenPos = n / 2; // total odd positions int oddPos = n - evenPos; int [] tempArr = new int [n]; // copy original array in an // auxiliary array for ( int i = 0; i < n; i++) tempArr[i] = arr[i]; // sort the auxiliary array Array.Sort(tempArr); int j = oddPos - 1; // Fill up odd position in // original array for ( int i = 0; i < n; i += 2) { arr[i] = tempArr[j]; j--; } j = oddPos; // Fill up even positions in // original array for ( int i = 1; i < n; i += 2) { arr[i] = tempArr[j]; j++; } // display array for ( int i = 0; i < n; i++) Console.Write(arr[i] + " " ); } // Driver Code public static void Main() { int [] arr = new int []{ 1, 2, 3, 4, 5, 6, 7 }; int size = 7; rearrangeArr(arr, size); } } /* This code is contributed by vt_m */ |
PHP
<?php // PHP program to rearrange the array // as per the given condition // function to rearrange the array function rearrangeArr(& $arr , $n ) { // total even positions $evenPos = intval ( $n / 2); // total odd positions $oddPos = $n - $evenPos ; $tempArr = array_fill (0, $n , NULL); // copy original array in an // auxiliary array for ( $i = 0; $i < $n ; $i ++) $tempArr [ $i ] = $arr [ $i ]; // sort the auxiliary array sort( $tempArr ); $j = $oddPos - 1; // fill up odd position in // original array for ( $i = 0; $i < $n ; $i += 2) { $arr [ $i ] = $tempArr [ $j ]; $j --; } $j = $oddPos ; // fill up even positions in // original array for ( $i = 1; $i < $n ; $i += 2) { $arr [ $i ] = $tempArr [ $j ]; $j ++; } // display array for ( $i = 0; $i < $n ; $i ++) echo $arr [ $i ] . " " ; } // Driver code $arr = array (1, 2, 3, 4, 5, 6, 7 ); $size = sizeof( $arr ); rearrangeArr( $arr , $size ); // This code is contributed // by ChitraNayal ?> |
Javascript
<script> // Javascript program to rearrange the array // as per the given condition // function to rearrange the array function rearrangeArr(arr, n) { // total even positions let evenPos = Math.floor(n / 2); // total odd positions let oddPos = n - evenPos; let tempArr = new Array(n); // copy original array in an // auxiliary array for (let i = 0; i < n; i++) tempArr[i] = arr[i]; // sort the auxiliary array tempArr.sort(); let j = oddPos - 1; // fill up odd position in original // array for (let i = 0; i < n; i += 2) { arr[i] = tempArr[j]; j--; } j = oddPos; // fill up even positions in original // array for (let i = 1; i < n; i += 2) { arr[i] = tempArr[j]; j++; } // display array for (let i = 0; i < n; i++) document.write(arr[i] + " " ); } // Driver code let arr = [ 1, 2, 3, 4, 5, 6, 7 ]; let size = arr.length; rearrangeArr(arr, size); //This code is contributed by Mayank Tyagi </script> |
Output:
4 5 3 6 2 7 1
Time Complexity: O( n logn )
Auxiliary Space: O(n)
Another Approach-
We can traverse the array the by defining two variables p and q and assign values from last.
if even index is there then we will give it max value other wise min value.
p =0 and q= end;
p will go ahead and q will decrease.
C++
#include <bits/stdc++.h> using namespace std; int main(){ int n,i,j,p,q; int a[]= {1, 2, 1, 4, 5, 6, 8, 8}; n= sizeof (a)/ sizeof (a[0]); int b[n]; for (i=0;i<n;i++) b[i]=a[i]; sort(b,b+n); p=0;q=n-1; for (i=n-1;i>=0;i--){ if (i%2!=0){ a[i]=b[q]; q--; } else { a[i]=b[p]; p++; } } for (i=0;i<n;i++){ cout<<a[i]<< " " ; } return 0; } |
Java
import java.util.*; class GFG{ public static void main(String[] args) { int n, i, j, p, q; int a[] = { 1 , 2 , 1 , 4 , 5 , 6 , 8 , 8 }; n = a.length; int []b = new int [n]; for (i = 0 ; i < n; i++) b[i] = a[i]; Arrays.sort(b); p = 0 ; q = n - 1 ; for (i = n - 1 ; i >= 0 ; i--) { if (i % 2 != 0 ) { a[i] = b[q]; q--; } else { a[i] = b[p]; p++; } } for (i = 0 ; i < n; i++) { System.out.print(a[i]+ " " ); } } } // This code is contributed by gauravrajput1 |
Python3
if __name__ = = '__main__' : #n, i, j, p, q; a = [ 1 , 2 , 1 , 4 , 5 , 6 , 8 , 8 ]; n = len (a); b = [ 0 ] * n; for i in range (n): b[i] = a[i]; b.sort(); p = 0 ; q = n - 1 ; for i in range (n - 1 , - 1 , - 1 ): if (i % 2 ! = 0 ): a[i] = b[q]; q - = 1 ; else : a[i] = b[p]; p + = 1 ; for i in range (n): print (a[i], end = " " ); # This code is contributed by gauravrajput1 |
C#
using System; public class GFG{ public static void Main(String[] args) { int n, i, j, p, q; int []a = {1, 2, 1, 4, 5, 6, 8, 8}; n = a.Length; int []b = new int [n]; for (i = 0; i < n; i++) b[i] = a[i]; Array.Sort(b); p = 0; q = n - 1; for (i = n - 1; i >= 0; i--) { if (i % 2 != 0) { a[i] = b[q]; q--; } else { a[i] = b[p]; p++; } } for (i = 0; i < n; i++) { Console.Write(a[i]+ " " ); } } } // This code is contributed by gauravrajput1 |
Javascript
<script> var n, i, j, p, q; var a = [ 1, 2, 1, 4, 5, 6, 8, 8 ]; n = a.length; var b = Array(n).fill(0); for (i = 0; i < n; i++) b[i] = a[i]; b.sort(); p = 0; q = n - 1; for (i = n - 1; i >= 0; i--) { if (i % 2 != 0) { a[i] = b[q]; q--; } else { a[i] = b[p]; p++; } } for (i = 0; i < n; i++) { document.write(a[i] + " " ); } // This code is contributed by gauravrajput1 </script> |
This algorithm will take 1 for loop less than the previous one.